Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $y = \dfrac{2n - 6}{-8n + 64} \times \dfrac{9n - 18}{n^2 - 5n + 6} $
Solution: First factor the quadratic. $y = \dfrac{2n - 6}{-8n + 64} \times \dfrac{9n - 18}{(n - 3)(n - 2)} $ Then factor out any other terms. $y = \dfrac{2(n - 3)}{-8(n - 8)} \times \dfrac{9(n - 2)}{(n - 3)(n - 2)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ 2(n - 3) \times 9(n - 2) } { -8(n - 8) \times (n - 3)(n - 2) } $ $y = \dfrac{ 18(n - 3)(n - 2)}{ -8(n - 8)(n - 3)(n - 2)} $ Notice that $(n - 2)$ and $(n - 3)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ 18\cancel{(n - 3)}(n - 2)}{ -8(n - 8)\cancel{(n - 3)}(n - 2)} $ We are dividing by $n - 3$ , so $n - 3 \neq 0$ Therefore, $n \neq 3$ $y = \dfrac{ 18\cancel{(n - 3)}\cancel{(n - 2)}}{ -8(n - 8)\cancel{(n - 3)}\cancel{(n - 2)}} $ We are dividing by $n - 2$ , so $n - 2 \neq 0$ Therefore, $n \neq 2$ $y = \dfrac{18}{-8(n - 8)} $ $y = \dfrac{-9}{4(n - 8)} ; \space n \neq 3 ; \space n \neq 2 $